Centre of Mass and Collision Part 1

The given situation can be represented as followsIn the above figure, XX′ be the axis which passes through A and is parallel to DB.
∴Moment of inertia of the system
= Mass × (Perpendicular distance from axis )2
I=m(AC)2+m(ED)2+m(FB)2
=m(0)2+m(I√2)2+m(‌

1

√2

)2+m(‌

1

√2

)2
=3ml2